Is every cauchy sequence bounded? I have understood the proof of this theorem but am confused in one example: $$X_n = \frac{1}{n-1}$$ This sequence is a cauchy sequence. I know that after $m$ terms the terms of the sequence falls into an open ball and is hence bounded but before those $m$ terms i.e. for $n=1$, $X_n$ is infinite.Then how can we say that every cauchy sequence is bounded? Any help is heartly appreciated.

For $n=1$ we have $n-1=0$ and so $\frac{1}{n-1}$ is not defined. So you cannot start your sequence at $n=0.$ $x_1$ is not infinite but $x_1$ is not defined, at least in the set of real numbers $\mathbb{R}$. The symbol $\infty$ is used in mathematics but you should always check what is its meaning in the context where it is used.

In the context you use it (a an element of the real numbers) it does absolutely make no sense and so you can not use it.

The sequence $$1,\frac12,\frac13,\ldots$$ (this is your sequence $x_2,x_3,x_4,\ldots$) is a Cauchy sequence and it is bounded. (What is a bound for this sequence?)

The sequences $1,2,3,4,\ldots$ and $1,2,1,2,1,2,1,2,\ldots$ are nto Cacuhy sequences but the second one is bounded the first one is not (Why?).

**Annotation**

One can construct extensions to the set of real numbers $\mathbb{R}$ that contain $\infty$ but statements that are valid in $\mathbb{R}$ must not be valid in this extenstion of $\mathbb{R}$